Radius and Center of a Circle from 3 Points

Date: 07/23/99 at 12:57:12
From: Nathan Sokalski
Subject: Radius and center of circle when given 3 points
I have been told during my past geometry classes that there is only
one circle that will go through any three given points. I understand
why and agree with this 100%. However, I am having trouble figuring
out what the center and radius are. Could you please tell me how to
calculate these two values?

Date: 07/24/99 at 15:51:38
From: Doctor Rob
Subject: Re: Radius and center of circle when given 3 points
Thanks for writing to Ask Dr. Math.
The equation of the circle given three points can be found in the Dr.
Math FAQ on analytic geometry formulas:
http://mathforum.org/dr.math/faq/formulas/faq.analygeom_2.html#twocircles
Expand the determinant, and complete the square on x and y to put it
into center-radius form. Then read off the center and radius.
Alternate approach: The center is the point equidistant from all three
points. Setting the squares of the distances equal will give you two
independent linear equations, which are the equations of the
perpendicular bisectors of two of the three line segments joining the
points. The place where those two lines intersect, that is, the common
solution of the two simultaneous linear equations, will be the center.
Then it is simple to compute the distance from the center to any of
the three points to get the radius.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/

Date: 07/26/99 at 01:05:48
From: Nathan Sokalski
Subject: Re: Radius and center of circle when given 3 points
When I went to the page you told me about, I did not understand what I
was supposed to do with the numbers. I assumed that the x and y (first
line) were the coordinates for the center, but what are the x, y, and
1 following the x^2+y^2 for? And I still don't understand exactly what
I am supposed to do with the numbers (maybe when I know what the x, y,
and 1 following the x^2+y^2 are for I'll figure that out, I don't
know). Thank you.
Nathan Sokalski

Date: 07/26/99 at 11:30:28
From: Doctor Rob
Subject: Re: Radius and center of circle when given 3 points
Sorry! I guess that you are not familiar with determinants. This
notation gives a compact way of writing the equation of the circle.
The less compact way gives you this equation:
(x^2+y^2-x3^2-y3^2)*(x1-x3)*(y2-y3)
+ (x1^2+y1^2-x3^2-y3^2)*(x2-x3)*(y-y3)
+ (x2^2+y2^2-x3^2-y3^2)*(x-x3)*(y1-y3)
- (x^2+y^2-x3^2-y3^2)*(x2-x3)*(y1-y3)
- (x1^2+y1^2-x3^2-y3^2)*(x-x3)*(y2-y3)
- (x2^2+y2^2-x3^2-y3^2)*(x1-x3)*(y-y3) = 0.
This equation is satisfied by all points whose coordinates are (x,y)
on the circle.
Expand all that out, and gather together all terms involving x^2, x,
y^2, and y. Then complete the square on x and y to put the equation in
the form
(x-h)^2 + (y-k)^2 = r^2.
Then the center of the circle is (h,k) and the radius is r.
Example: Suppose the points are (x1,y1) = (5,0), (x2,y2) = (3,3), and
(x3,y3) = (2,-4). Then the equation found by substituting in the above
is
(x^2+y^2-4-16)*(5-2)*(3+4) + (25+0-4-16)*(3-2)*(y+4)
+ (9+9-4-16)*(x-2)*(0+4) - (x^2+y^2-4-16)*(3-2)*(0+4)
- (25+0-4-16)*(x-2)*(3+4) - (9+9-4-16)*(5-2)*(y+4) = 0,
(21*x^2+21*y^2-420) + (5*y+20) + (-8*x+16) - (4*x^2+4*y^2-80)
- (35*x-70) - (-6*y-24) = 0,
17*x^2 + 17*y^2 - 43*x + 11*y - 210 = 0,
x^2 + y^2 + (-43/17)*x + (11/17)*y = 210/17,
x^2 - (43/17)*x + 43^2/34^2 + y^2 + (11/17)*y + 11^2/34^2
= (210*68+43^2+11^2)/34^2,
(x - 43/34)^2 + (y + 11/17)^2 = 16250/34^2 = (25*sqrt[26]/34)^2.
The center is (43/34, -11/17) and the radius is 25*sqrt[26]/34.
I suppose one could do this entire calculation with the equation first
above, and get complicated formulas involving x1, y1, x2, y2, x3, and
y3 which would tell you the center (h,k) and radius r, but I don't
think I want to do the algebra. Furthermore, they would be hard to
memorize.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/

Date: 07/26/99 at 15:35:08
From: Nathan Sokalski
Subject: Re: Radius and center of circle when given 3 points
I understand the plugging in of the x1, y1, x2, y2, x3, and y3, and I
understand how to get to the step
x^2 + y^2 + (-43/17)*x + (11/17)*y = 210/17,
but I was confused while moving to the next step
x^2 - (43/17)*x + 43^2/34^2 + y^2 + (11/17)*y + 11^2/34^2
= (210*68+43^2+11^2)/34^2
and the steps after that. I am confused as to where some of the
numbers such as 34 came from. If possible, could you please describe
these last few steps to me with a little bit more detail? Thank you.
Nathan Sokalski

Date: 08/03/99 at 22:32:09
From: Doctor Rob
Subject: Re: Radius and center of circle when given 3 points
See this answer in the Dr. Math archives for tips on "Completing the
Square":
http://mathforum.org/dr.math/problems/robertd5.22.96.html
43^2/34^2 is the square of half the x-coefficient, -43/17.
11^2/34^2 is the square of half the y-coefficient, 11/17. This works
because, for any expressions a and z,
z^2 + a*z = (z+a/2)^2 - (a/2)^2.
Apply this with z = x, a = -43/17, and add (a/2)^2 to both sides. Then
apply it again with z = y and a = 11/17, and add (a/2)^2 to both
sides.
Then it is just a matter of doing the arithmetic.
- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/